You have misunderstood how "high voltage" is beneficial, in the context of conduction losses.
The first part of this answer doesn't not address your question directly, but it will help you must understand why "high voltage" can help, and the context in which this principle is valid.
In power delivery systems, loss occurs in the conductors themselves, due to electrical resistance of those conductors. As you know, Ohm's law says that any conductor will develop a potential difference between its two ends proportional to the current flowing through the conductor, due to its resistance. Here's the scenario:
![schematic](https://cdn.statically.io/img/i.sstatic.net/CPBwK.png)
simulate this circuit – Schematic created using CircuitLab
RW1 and RW2 represent resistance of the wires carrying current to and from the load, represented by RL. Current is 1A, the same everywhere in the loop, but wire resistance develops a total of 2V, for a dissipation of \$2 \times 1A \times 1V = 2W\$ of power, lost in the wires.
The load receives only 10V of the total 12V available from source V1, after the wires have "eaten" 2 of them, and therefore dissipates \$1A \times 10V = 10W\$.
Now we'll increase the supply voltage V1 by a factor of 10, to become 120V, and adjust RL so that it still consumes only 10W:
![schematic](https://cdn.statically.io/img/i.sstatic.net/LU9Ez.png)
simulate this circuit
Now we have only 14mW of power lost in the wires (a factor of 140 improvement), and still 10W being delivered to the load. It may seem contrived that we had to change the load resistance, but the point here is that we are delivering the same power to the load, but with vastly reduced conduction losses in the wires, and that reduced loss is due to the reduction of loop current that we asking the wires to carry.
This principle is used in all national power grids, to minimise power dissipation in the wires, and reduce the diameter of wires needed to carry current. While we don't directly "change the resistance" of the load (that's determined by the households, factories and whatever else is connected), an equivalent function is performed by transformers.
It's important to understand that voltage of the the 12V or 120V source isn't "travelling through" anything. It's being shared between whatever elements exist in the loop. It's the current that's flowing through things, in combination with the elements' resistance, that is causing power to be delivered to them.
Now back to your energy storage system. In such a system, a zero-resistance coil has a current passed through it, which develops a magnetic field in which the energy is stored. Current is allowed to continue flowing around the loop, and the absence of resistance in the loop means that no power is delivered to anything. By conservation of energy, the energy of the magnetic field is preserved.
To retrieve that energy, a resistance (or more generally, an "impedance") is introduced into the loop. This new element does not have zero impedance, and must therefore develop a voltage across it, by Ohm's law, which means that it receives \$P = I \times V\$ Joules of energy per second. That energy comes from the magnetic field.
Evidently, any resistance in the loop will receive power, causing the magnetic field to diminish over time, until it's completely exhausted, whether this is during "charging" of the storage system, or during "discharge". If the coils themselves have any resistance, they will dissipate power, and drain the energy of the magnetic field, by heating.
This is why it's necessary to use super-conducting coils; the loss of energy to non-zero resistances in the loop, including the coil itself, will occur regardless of the voltages present anywhere in the system. The success of such an energy storage system depends entirely on the resistances present, and not on what voltage you use to "charge up" the store.
In fact, since the idea is to establish a "semi-permanent" current in the loop, it doesn't make sense to talk of reducing current there. The current that flows is necessarily as high as you can make it, because it's intimately linked to the strength of magnetic field.
I hope you can see why the principle of using "high voltage" to improve efficiency is not applicable in this scenario, and it doesn't matter what voltage you use to "charge up" the energy store. The goals for the storage system are chiefly:
Develop a strong magnetic field, which necessitates high current
Have zero (or as close to zero as is possible) resistance in the current loop, to avoid depleting the energy of the field
Neither of these are going to be helped by using high voltages. As a further talking point, this might surprise you; when there's nothing receiving energy in the loop (no resistance present), the voltage across the coils must be zero, even when thousands of amps are flowing.